path independance

[CPL.Luke]

so, I'm having a hard time with path independance, namely with how do you find the scalar function that your vector function is the gradient of.

 

 

or G in

 

(nabla)G = f

[swansont]

Integrate it

[CPL.Luke]

alright then I suppose my question lies more in what happens to the i's the j's and the k's when you integrate,

 

for instance when you take the integral of f to get the scalar function G

[Tom Mattson]

alright then I suppose my question lies more in what happens to the i's the j's and the k's

 

You leave them out of it. If [imath]\vec{\nabla}G=\vec{f}[/imath] then:

 

[math]\frac{\partial G}{\partial x}=f_x[/math]

 

[math]\frac{\partial G}{\partial y}=f_y[/math]

 

[math]\frac{\partial G}{\partial y}=f_z[/math]

[CPL.Luke]

then in the end you just do fx+fy+fz correct? without an ij or k making it a scalar function.

[Tom Mattson]

No, in the end you have f(x,y,z).